Question

An AC power generator produces 50 A (rms) at 3600 V. The voltage is stepped up to 100 000 V by an ideal transformer and the energy is transmitted through a long-distance power line which has a resistance of 100 ohms. What percentage of the power delivered by the generator is dissipated as heat in the long-distance power line?

Answer 1

Given:

`I_(rms) = 50 A`

voltage, V = 3600V

step-up voltage, V' = 100000 V

Resistance of line,
`R = 100\ohm`

Solution:

To calculate % heat loss in long distance power line:

Power produced by AC generator, P =
`50* 3600` W

P = 180000 W = 180 kW

At step-up voltage, V = 100000V or 100 kV

current, I =
`(P)/(V')`

I =
`(1800000)/(100000)`

I = 1.8 A

Power line voltage drop is given by:

`V_(drop) = I* R`

`V_(drop) = 1.8* 100`

`V_(drop) = 180 V`

Power dissipated in long transmission line
`P_(dissipated) = V_(drop)* I`

Power dissipated in long transmission line
`P_(dissipated) = 180* 1.8` = 324 W

% Heat loss in power line,
`P_(loss) = (P_(dissipated))/(P)* 100`

% Heat loss in power line,
`P_(loss) = (324)/(180000)* 100`

`P_(loss) = 0.18%`